2
$\begingroup$

All rings are assumed to have unity: $1\in R$.

In a ring $R$:

Definition: An element is a unit if it is invertible (it has a 2 sided multiplicative inverse).


In a commutative ring $R$:

Definition: An ideal $P \ne R$ is a prime ideal if $ab \in P \implies a\in P \text{ or } b \in P$

Definition: An element $p\in R$ is a prime element if $(p)$ is a prime ideal.


In an integral domain $R$:

Definition: For $a,b \in R$, we say $a$ is a factor or divisor of $b$ if $b\in (a)$.

Definition: For $a,b \in R$, we say $a$ is a proper factor of $b$ if $b \in (a)$ and $a \notin (b)$.

Definition: An element $a\in R$ is irreducible if it is not a unit and has no proper factors other than units.


Known:

In an integral domain:
$(p)$ is prime $\iff$ $p$ is prime $\implies$ $p$ is irreducible.

In a unique factorization domain:
$(p)$ is prime $\iff$ $p$ is prime $\iff$ $p$ is irreducible.


Question:

If I have a prime ideal, that is not principal (but finitely generated), say $(p,q)$, what can I conclude about $p$ and $q$? Can I say they are prime? In an integral domain, can I say they are irreducible? What additional criteria are needed to make these conclusions?

What about the other way, if $p$ and $q$ are prime, then can I conclude $(p,q)$ is a prime ideal? What additional criteria are needed?

I'm also interested in extending these ideas outside of domains and to noncommutative rings, but it's more important to understand them in an integral domain first.

$\endgroup$
1
  • 2
    $\begingroup$ In the first case, no, you can't say that $p.q$ are prime. First of all, there isn't one set of generators. If $p,q$ are generators, then so are $p+aq,q$ for any $a\in R$. If you were correct that $p,q$ must be prime, then you must have $p+aq$ must always be prime. When $a$ is a multiple of $p$, this means that $1+bq$ must always be a unit. Taking $R=\mathbb Z[x]$, $p=2+2x,q=x$ ,then $(p,q)=(2+2x,x)=(2,x)$ is prime, but $2+2x$ is not prime. $\endgroup$ Nov 11, 2015 at 17:48

1 Answer 1

3
$\begingroup$

I'm skipping the questions about irreducibles. Too many questions here.

If you take the typical example: $R=\mathbb Z[\sqrt{-5}]$ where:

$$2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\tag{1}$$

Then the ideal, $I=(2,1+\sqrt{-5})$ is not principle, and it is prime - it is easy to show that $R/I\cong \mathbb Z/2\mathbb Z$, which is a field, so $I$ is maximal, hence prime.

You also have that $2,3,1\pm \sqrt{-5}$ are irreducible, but not prime.

It is also easy to show that no element of $I$ is prime, since you can show that $R/(p)$ is finite for any $p\in I$, and if $(p)$ is prime, $R/(p)$ is an integral domain, and a finite integral domain is a field. Thus $(p)$ is maximal. But $(p)\subsetneq I\subsetneq R$, so $(p)$ is not maximal.

So there are times where you can't find any prime element of a prime ideal.

In $\mathbb Z[x]$, $x^2+1$ and $5$ are prime, but $(x-2)(x+2)=(x^2+1)-5\in (5,x^2+1)$. So, no, if $p,q$ are prime, then it is not necessarily true that $(p,q)$ is prime.

According to Wikipedia, in UFDs, every irreducible element is prime (which practically goes without saying,) but not visa-versa. The converse is true when the ring satisfies the "ascending chain condition for principle ideals."

$\endgroup$
8
  • $\begingroup$ Thanks, I've worked out everything except why $R/(p)$ must be finite for any $p \in I$. Why is that? $\endgroup$ Nov 12, 2015 at 1:07
  • $\begingroup$ Clearly my questions (about multiple generators) only apply outside a PID. I know that this example $\mathbb Z [\sqrt{-5}]$ is an ID, but not a UFD. I'm now wondering if this relies on the multiple ways to factor ... if this still holds in a UFD where prime and irreducible coincide. $\endgroup$ Nov 12, 2015 at 1:10
  • 1
    $\begingroup$ It's true for any $p\in R$, actually, and it is because every element $p\in R$ is a divisor of an integer $n$, and thus the set $a+b\sqrt{-5}$ with $0\leq a,b<n$ covers all elements of $R/(p)$. So there are at most $n^2$ elements of $R/(p)$. $\endgroup$ Nov 12, 2015 at 1:16
  • 1
    $\begingroup$ Other than $\mathbb Z[x]$ and the like, I don't know many UFDs that aren't PIDs, but the key is that all of ours assumptions about what it means to be irreducible and prime for elements of a general ring/integral domain go out the window. It is worth knowing why the word "ideal" was chosen. Originally. they were called "ideal divisors." As in: Ideally, there should exist a GCD of $2$ and $1+\sqrt{-5}$. Drats! Why isn't there one? Well, let's pretend there is one! We'll call it an "ideal divisor."It wasn't until later that the abstraction was formalized. $\endgroup$ Nov 12, 2015 at 13:45
  • $\begingroup$ According to wikipedia, in UFDs, every irreducible element is prime, but not visa-versa. The converse is true when the ring satisfies the "ascending chain condition for principle ideals." $\endgroup$ Nov 12, 2015 at 14:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .